The infinite sum of integral of positive function is bounded so function tends to 0

Question

Let $f_n(x)$ be positive measurable functions such that $$\sum_{n=1}^\infty \int f_n \lt \infty.$$ Show that $f_n \to 0$ almost everywhere.

Attempt:

Let $\displaystyle K = \sum_{n=1}^\infty\int f_n$ and $\displaystyle S_m = \sum_{n=1}^m \int f_n$. Then, $\forall \epsilon \gt 0$, $\exists L$ such that $\forall m \gt L$, $|S_m - L| \le \epsilon$.

That is, $\displaystyle \sum_{n=m+1}^\infty \int f_n \lt \epsilon$. Therefore, $\forall n \gt L$ we have $\displaystyle \int f_n \lt \epsilon $, then the result should follow.

I don't know why the grader of my class said this proof is wrong.

If I am truly wrong, where is my error?

Thanks!

Answer 1

By the Monotone Convergence Theorem, $$\sum_{n=1}^{\infty}\int f_{n}\ d\mu = \int \sum_{n=1}^{\infty} fn\ d\mu < \infty$$ The above implies that $\sum_{n=1}^{\infty}f_{n} < \infty$ a.e. $(\mu)$. Therefore $f_{n}(x) \rightarrow 0$ a.e. ($\mu$).

Answer 2

Just because a sequence of functions has integral tending to zero doesn't give you pointwise convergence. For example, one standard example is to take a sequence of indicator functions moving back and forth inside $[0,1]$ that have smaller and smaller support; for example, indicators on $[0, 1/2]$ and $[1/2, 1]$< then $[0, 1/3], [1/3, 2/3]$ and $[2/3, 1]$, and so on. The best result you can get is a subsequence which converges pointwise a.e.

For a start in a different direction that does lead to a proof, consider applying the Borel-Cantelli lemma.

Answer 3

Define $$ E_k=\left\{x:\limsup_{n\to\infty}f_n(x)\ge\frac1k\right\} $$ For each $x\in E_k$, $f_n(x)\ge\frac1{2k}$ infinitely often. Therefore, for each each $x\in E_k$ $$ \sum_{n=1}^\infty f_n(x)=\infty $$ Thus, if the measure of $E_k$ is positive, $$ \int_{E_k}\sum_{n=1}^\infty f_n(x)\,\mathrm{d}x=\infty $$ Therefore, the measure of each $E_k$ must be $0$. Thus, $$ \left|\bigcup_{k=1}^\infty E_k\right|\le\sum_{k=1}^\infty\left|E_k\right|=0 $$ However, $$ \left\{x:\limsup_{n\to\infty}f_n(x)\ne0\right\}\subset\bigcup_{k=1}^\infty E_k $$ Therefore, for almost every $x$, $$ \lim_{n\to\infty}f_n(x)=0 $$

Answer 4

I'm going to assume

1. $f_n$'s are $\mu$-integrable and in measure space $(S, \Sigma, \mu)$.

2. By $\int X$ you mean $\int_S X d\mu$

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Since the $f_n$'s are nonnegative because they are positive, we can switch summation and integral (this is because of MCT):

$$\sum_{n=1}^{\infty}\int_S f_n d\mu = \int_S \sum_{n=1}^{\infty} f_n d\mu < \infty$$

I recall a rule in probability theory that states $E[X] < \infty \to X < \infty \ \mu-\text{a.s.}$ (to see why, try to prove the contrapositive). I'm guessing this comes from measure theory: $\int_S X d \mu < \infty \to X < \infty \ \mu-\text{a.e.}$

Hence we have that $\sum_{n=1}^{\infty}f_{n} < \infty \ \mu-\text{a.e.}$.

We know from basic calculus that $\sum_n a_n < \infty \to \lim a_n = 0$

$a_n$ is a sequence of numbers, but it applies to a sequence of functions, something we know from real analysis.

$\therefore, \lim f_n = 0 \ \mu-\text{a.e.}$.